Not approximately, but precisely, and what does one need to know to be able to do that?
For instance, I know that $\tan(45°)$ gives the incline of a line which encloses $45°$ with the $x$-axis, namely $1$, therefore $\arctan(1)=45°$ or $\frac{\pi}{4}$. Is there a way to use this relation?
Regards
Take a 30-60-90 triangle, with sides $1,\sqrt{3},2$. Extend the side of length $\sqrt{3}$ by an extra two units. That gives a $15-75-90$ triangle whose short sides are 1 and $2+\sqrt{3}$
Well, you can go the other way. That's easier.
$\tan 30^{\circ} = \frac{1}{\sqrt 3}$
Let $t = \tan 15^{\circ}$.
So $\displaystyle \frac{2t}{1-t^2} = \frac{1}{\sqrt 3}$
Solving,
$t^2 + 2\sqrt 3 t - 1 = 0$
$\displaystyle t = -\frac{-2\sqrt 3 + \sqrt{12+4}}{2}$ (only admissible root)
$t = 2 -\sqrt 3$
(edited as the question seems to have been amended to ask for $\tan (-15^{\circ})$:
$\tan (-15^{\circ}) = -\tan 15^{\circ} = \sqrt 3 - 2$
You can use $$ \tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta} $$ with $\alpha=45^\circ$ and $\beta=60^\circ$, so $\tan45^\circ=1$ and $\tan60^\circ=\sqrt{3}$: $$ \tan(-15^\circ)=\frac{1-\sqrt{3}}{1+\sqrt{3}}=\frac{(1-\sqrt{3})^2}{1-3} =\frac{1-2\sqrt{3}+3}{-2}=\sqrt{3}-2 $$
Here is yet another way. $$2-\sqrt{3}=\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{2}=\frac{\sqrt{3}-1}{1+\sqrt{3}}$$
Now we have $$\arctan \left(\frac{a-b}{1+ab}\right)=\arctan a-\arctan b$$
So $$\arctan 2-\sqrt{3}=\arctan \sqrt{3}-\arctan 1=60^{\circ}-45^{\circ}=15^{\circ}.$$
The angle occurs in the dodecagon, which is a composition of triangles and squares. This gives you a coordinate system, from which ye find
sin(15) = (sqrt (3)+1)/sqrt(2) cos(15) = (sqrt (3)-1)/sqrt(2)
Whence the ratio is (sqrt(3)-1)̃²/(sqrt(3)+1)(sqrt(3)-1) = 2-sqrt(3).
$$\sqrt3-2=\cot30^\circ-\csc30^\circ=\dfrac{\cos30^\circ-1}{\sin30^\circ}=\dfrac{-2\sin^215^\circ}{2\sin15^\circ\cos15^\circ}=-\tan15^\circ=\tan(-15^\circ)$$
Now use $-90^\circ<\arctan x<90^\circ$
